WorkWork

F

The force, F, pushes the box for a short

distance. This causes the box to start

moving!!!!!!It gains energy!!!!

I just don’t have any energy Whoa!!! Now I do!!!

WorkWorkIf a force, F, is used to move an object a displacement, d, then the force does work on

the object.

F

d

W = Fd

Units for WorkUnits for Work The units for force are Newtons (N) and The units for force are Newtons (N) and

the units for displacement are meters the units for displacement are meters (m).(m).

Since Work is equal to force times Since Work is equal to force times displacment. We could use a Newton-displacment. We could use a Newton-meter (Nmeter (N··m).m).

But the more common unit is the Joule (J)But the more common unit is the Joule (J) 1 J = 1 N1 J = 1 N··mm 1 J is defined as the amount of work for 1 J is defined as the amount of work for

1- N of force to move an object 1-m1- N of force to move an object 1-m

In order for work to be done:In order for work to be done:

There must be a force There must be a force applied.applied.

The object must have a The object must have a displacement.displacement.

The force must be responsible The force must be responsible for the displacement.for the displacement.

How much work is being done How much work is being done by this weightlifter?by this weightlifter?

NONE!!!!!

Since he doesn’t move the

weights any distance (There

is no displacement),

the work done is 0 J.

How much work is being done How much work is being done in these cases?in these cases?

NONE!!!!!In each case the forces

are perpendicular to the displacement,

therefore they are not “responsible” for the

motion.

In general, when a force is perpendicular

to the motion of an object, it DOES NO

WORK!!!!

Watch Video

The weight lifter in the video lifts 263.5-kg a distance of 2.43-m. How much work does he do during one lift?

If he lifts it twice, how much work would he do?

F

df

If more than one force is acting on an object, you can find the total work done on the object

by adding the work done by each force together.

If the force is in the same direction as the motion, then the work done is positive because

it adds energy to the object.

If the force is in the opposite direction as the motion, then the work done is negative because

it removes energy from the object.

A box is being pushed with a force of 25-N. There is also friction between the box and the table of 12-N. The box is moved 3.0-m

How much work is being done by the 25-N force?

How much work is being done by the friction?

How much total work is being done on the box?

F = 25-N

d = 3.0-m

W25 = Fd = (25)(3.0) = 75-J

This work is positive because the force is in the direction of the motion and adds energy to the

box.

25-N

3.0-m12-N

W12 = -fd = -(12)(3.0) = -36-J

This work is negative because the force is in the opposite direction of the motion and removes

energy from the box.

d = 3.0-m

f = 12-N

A box is being pushed with a force of 25-N. There is also friction between the box and the table of 12-N. The box is moved 3.0-m. Find the total work done on the box.

What is the net force on the box?

How much total work is being done on the box?

F = 25-N

25-N

3.0-m

12-N

d = 3.0-m

f = 12-N

SAME PROBLEM, DIFFERENT METHODSAME PROBLEM, DIFFERENT METHOD

Fnet = F – f = 25 – 12 = 13-N

Fnet = 13-N

Another way to find the total work done on the box is to find

the net force on the box and multiply this by the

displacement.